Question

If $G(x)$ be the distribution function of random variable $X$ symmetric about $0$ then $\int_{-a}^{a} G(x)dx$ equals

• A. $a$
• B. $2a$
• C. $0$
• D. $1$

Hint:

For any CDF $F(x)$ symmetric about $0$, the area under the curve from $-a$ to $a$ is always equal to $a$. This is a very useful shortcut in probability theory problems involving integration of symmetric CDFs.

Answer 1

The Correct Option is A

To evaluate the integral of the distribution function for a symmetric random variable, we utilize the specific properties of symmetry related to the cumulative distribution function (CDF).

Step 1: \color{redUtilizing the Property of Symmetry
For a random variable $X$ that is symmetric about $0$, the distribution function $G(x)$ satisfies the following identity:
$G(-x) = 1 - G(x)$ for all $x$.
This is derived from the fact that for a symmetric distribution about zero, $P(X \le -x) = P(X \ge x) = 1 - P(X < x)$.

Step 2: \color{redSplitting the Integral
We are required to find the value of $I = \int_{-a}^{a} G(x) dx$.
We can split the limits of integration from $-a$ to $0$ and from $0$ to $a$:
$I = \int_{-a}^{0} G(x) dx + \int_{0}^{a} G(x) dx$

Step 3: \color{redVariable Substitution for the First Part
In the first integral $\int_{-a}^{0} G(x) dx$, let us substitute $x = -u$, then $dx = -du$.
As $x$ goes from $-a$ to $0$, $u$ goes from $a$ to $0$:
$\int_{a}^{0} G(-u) (-du) = \int_{0}^{a} G(-u) du$
Using the symmetry property $G(-u) = 1 - G(u)$, the integral becomes:
$\int_{0}^{a} (1 - G(u)) du = \int_{0}^{a} 1 du - \int_{0}^{a} G(u) du$

Step 4: \color{redCombining the Results
Now substitute this back into our original split integral $I$:
$I = [ \int_{0}^{a} 1 du - \int_{0}^{a} G(u) du ] + \int_{0}^{a} G(x) dx$
The terms $\int G(u) du$ and $\int G(x) dx$ are identical and cancel each other out:
$I = \int_{0}^{a} 1 du = [u]_{0}^{a} = a$
Thus, the integral equals $a$.